Question

- 16(x - 2)² – 25=0​

Answer 1

Answer:

x=2±16i$\sqrt{73}$

Step-by-step explanation:

1.) Using the PEMDAS distribute the $(x-2)^{2}$ using the FOIL method

$(x-2)^2= x^2-4x+4$

2.) Distribute the -16 to ($(x-2)^2= x^2-4x+4$)

$-16(x^2-4x+4)= -16x^2+64x-64$

3.) Add the -25 to the equation

$(-16x^2+64x-64)-25=-16x^2+64x-89$=0

4.) Since this isn't factor-able, you must use the quadratic formula,

$\frac{(-b (+ or -) \sqrt{(b^2-4ac)}}{2a}$ and distrubute the numbers

a=-16 b=64 and c=-89

5.) Simplify the radical by finding the factors of -18688 that has a perfect root as one of the factors (which is -256 and 73).

$\sqrt{-18688} =\sqrt{-256*73} =16i\sqrt{73}$

6.)Simplify the fraction

$\frac{-64(+ or -)16i\sqrt{73} }{-32} ={2}(+ or -)16i\sqrt{73}$

7.) Your two roots are:

{2}(+ or -)16i\sqrt{73}